2 /* function search: element searching in a sorted list in logarithmic complexity
3 ran 100000000 times on 3 lists of size 100 */
7 {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21,
8 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40,
9 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59,
10 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78,
11 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97,
14 {99, 98, 97, 96, 95, 94, 93, 92, 91, 90, 89, 88, 87, 86, 85, 84, 83, 82, 81,
15 80, 79, 78, 77, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62,
16 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43,
17 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24,
18 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3,
21 int search (int tab[], int e) {
22 int i, size, left, right, previous, middle;
26 if (size == 0) return 0;
30 middle = (left + right) / 2;
31 while ((middle != left) && (middle != right)) {
32 if (e == tab[middle]) return 1;
33 if (e > tab[middle]) left = middle;
35 middle = (left + right) / 2;
42 int i, search1, search2, search3;
44 {-21, 19, -26, -62, -72, 33, -14, 20, 85, 92, -26, -33, 38, -14, -88, -35,
45 -51, 29, -84, -98, -92, -60, -52, 53, 53, -11, -99, 72, -55, 90, -83, 98,
46 29, -25, -32, -33, -98, 14, -86, 99, 77, -38, -1, -7, -98, 57, 5, -64, 72,
47 60, -5, -66, -7, -84, -97, 92, 2, 11, 70, -27, -56, -16, -81, 86, 27, 24,
48 68, -54, -15, -45, 60, 25, -31, 67, 7, -14, -41, -67, 51, -27, -1, -98,
49 -97, -16, -41, 33, -8, -45, 48, -44, -1, 56, -70, 59, -60, 64, 5, 32, 86,
52 for (i = 0 ; i < 100000000 ; i++) {
53 search1 = search(tab1, 18);
54 search2 = search(tab2, -1);
55 search3 = search(tab3, 36);